AndreG wrote:
It takes printer A 4 more minutes than printer B to print 40 pages. If working together, the two printers can print 50 pages in 6 minutes, how long will it take printer A to print 80 pages?
(C) 2008 GMAT Club - m17#20
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I can get up to the part where:
\(\frac{1}{A} + \frac{1}{A - 4} = \frac{5}{24}\)
How do I solve this equation?? I wind up with some pretty complicated squared equation...
Thanks!
Even though it has been discussed before, and there is nothing to add to the solution, there is this one thing, which might just help.Apologies for the lack of brevity.
Imagine a quadratic which reads as : \(\frac {1}{x} + \frac{1}{x+1} =\frac{5}{6}\)
Try calculating this quadratic, it would take atleast say 1 minute. Now, take another look at the quadratic. We see that the RHS has 6 in the denominator. Thus, the LCM of both x and (x+1) should be 6.
I say
should be, because the RHS denotes a reduced fraction.
Now, think of the first two numbers which have an LCM of 6,and to top it up, they are consecutive \(\to\) 2 and 3[
x and (x+1)]
\(\frac {1}{x} + \frac{1}{x+1}\) = \(\frac {1}{2} + \frac{1}{3} =\frac{2+3}{6}\) = \(\frac{5}{6}\)
x = 2.
Also, as because there is only ONE correct positive answer for such problems, you can afford the luxury to neglect the second root, which in most cases would turn out to be a negative entity.
Now back to the current problem :
\(\frac{1}{A} + \frac{1}{A - 4} = \frac{5}{24}\)
Doesn't take long to figure out that the RHS is again a reduced fraction. Now, juggle between numbers which might give an LCM of 24 : (24,1),(6,8),(12,8),etc. But, notice that along with the LCM, the pair also subscribes to another restraint: they are seperated by 4 units. One can quickly recognize that (12,8) is one such pair, and this indeed gives us \(\frac{5}{24}\), when added.
Another way to look at it is by redistributing the numerator (5,in this case) into 2 parts, so that each part, is a factor of the common denominator(24), i.e. 5 \(\to\) (1,4) or (2,3). The first pair will lead to this \(\to \frac{1+4}{24}\) = \(\frac {1}{24} + \frac{1}{6}\) and we know that this is not a valid solution. Trying the other pair, we have something like this \(\to \frac{2+3}{24}\)= \(\frac {2}{24} + \frac{3}{24}\) = \(\frac {1}{8} + \frac{1}{12}\) and this is the correct solution.
Just to get the hang of it, let's try doing another problem :
Say \(\frac{1}{A} - \frac{1}{A+3} = \frac{1}{36}\).
LCM of 36\(\to\) (1,36),(4,9),(9,12),etc. But one pair which is seperated by 3 units \(\to\) (12,9) and it is the answer.
Another way, by redifining the numerator\(\to\) As the numertor in this case is anyways unity, we can think of it as the difference of 2 integers, both being factors of 36 \(\to\) (3,4).
Thus, \(\frac{1}{36} = \frac{4-3}{36} = \frac{4}{36} - \frac{3}{36} = \frac{1}{9} - \frac{1}{12}.\)
Note: This method is not fail-proof, but knowing that the guys at GMAC design excellent problems,I am sure this method might just come in handy in some way or the other.
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All that is equal and not-Deep Dive In-equality
Hit and Trial for Integral Solutions